Precalc: find a^3 when a = 8(cos(7pi/12) + i*sin(7pi/12))

[fraiha]

I'm doing a precalculous review for my class and while I've been able to figure out most of the problems using help from my textbook and the internet there are a few that I am unsure of and was hoping someone here could help me.


1) Find a^3 when a = 8(cos(7pi/12) + i*sin(7pi/12)
I'm sorry for this but I really have no idea where to begin with this. If anyone could provide some insight of a website on how to figure complex numbers out I would be appreciative.

2) Change the equation to rectangular and determine which conic section is represented by:
r = 5/(4+3cos(Theta))
If x = rcos(Theta) then x = (cos(Theta)*5)/(4+3cos(Theta) but I fail to see anything past this point and with that what the conic section is.

I'd appreciate any help on either of the problems and I apologize that I couldn't provide any insight myself. Thanks in advance!

 

[galactus]

Re: Precalc: find a^3 when a = 8(cos(7pi/12) + i*sin(7pi/12)

2) Change the equation to rectangular and determine which conic section is represented by:
r = 5/(4+3cos(Theta))
If x = rcos(Theta) then x = (cos(Theta)*5)/(4+3cos(Theta) but I fail to see anything past this point and with that what the conic section is.

Hello fraiha:

Remember, \(\displaystyle \L\\r=\sqrt{x^{2}+y^{2}}\)

You have: \(\displaystyle \L\\\sqrt{x^{2}+y^{2}}=\frac{5}{4+3(\frac{x}{\sqrt{x^{2}+y^{2}}})}\)

Solve for y:

\(\displaystyle \L\\4\sqrt{x^{2}+y^{2}}+3x=5\)

\(\displaystyle \L\\\sqrt{x^{2}+y^{2}}=\frac{5-3x}{4}\)

\(\displaystyle \L\\x^{2}+y^{2}=\frac{5-3x}{4}\)

\(\displaystyle \L\\y^{2}=(\frac{5-3x}{4})^{2}-x^{2}\)

\(\displaystyle \L\\y^{2}=\frac{-\7}{16}x^{2}-\frac{15}{8}x+\frac{25}{16}\)

\(\displaystyle \L\\y=\pm\frac{\sqrt{-7x^{2}-30x+25}}{4}\)

\(\displaystyle \L\\16y^{2}=-7x^{2}-30x+25\)

\(\displaystyle \L\\16y^{2}+7x^{2}+30x-25=0\)

Know what this represents?.

This is probably more than is necessary, but I thought the algebra practice wouldn't hurt.

 

[fraiha]

Hey thanks for the help, but I'm having problems understanding what you did.

Namely the part where cosX = X/sqrt(X^2+Y^2) ??. What logic am I missing here? But I do understand where you are going with this as now that I have the basic idea i'm to work my way through.

Ah and the conic would be an ellipse.

Again, I appreciate your help and I'm sorry for the trouble.

edit - I found out the logic i was missing, again you helped me quite a bit.

 

[soroban]

Re: Precalc: find a^3 when a = 8(cos(7pi/12) + i*sin(7pi/12)

Hello, fraiha!

1) Find \(\displaystyle a^3\) when \(\displaystyle a \:= \:8\left(\cos\frac{7\pi}{12}\,+\,i\sin\frac{7\p}{12}\right)\)

You're expected to be familiar with DeMoivre's Theorem:
\(\displaystyle \;\;\;(\cos\theta\,+\,i\sin\theta)^n\;=\;\cos(n\theta)\,+\,i\sin(n\theta)\)

Isn't that amazing? . . . The exponent jumps in and become a coefficient of \(\displaystyle \theta.\)


Your problem has: \(\displaystyle a\;=\;8\left(\cos\frac{7\pi}{12}\,+\,i\sin\frac{7\pi}{12}\right)\)

Cube both sides: \(\displaystyle \,a^3\;=\;\left[8\left(\cos\frac{7\pi}{12}\,+\,i\sin\frac{7\pi}{12}\right)\right]^3\)

Then we have: \(\displaystyle \,8^3\left[\cos\left(3\cdot\frac{7\pi}{12}\right)\,+\,i\sin\left(3\cdot\frac{7\pi}{12}\right)\right] \;= \;512\left(\cos\frac{7\pi}{4}\,+\,i\sin\frac{7\pi}{4}\right)\)

\(\displaystyle \;\;\;=\;512\left(-\frac{sqrt{2}}{2}\,+\,i\frac{\sqrt{2}}{2}\right) \;= \;256\sqrt{2}(-1\,+\,i)\)

2) Write \(\displaystyle \,r\:=\:\frac{5}{4\,+\,3\cos\theta}\,\) in rectangular coordintes.

We have: \(\displaystyle \,r(4\,+\,3\cos\theta)\;=\;5\)

Then we have: \(\displaystyle \;4r\,+\,3\underbrace{r\cos\theta}\;=\;5\)
. . . . . . . . . . . . . .\(\displaystyle \downarrow\;\;\;\;\;\downarrow\)
\(\displaystyle \;\:\)Hence: \(\displaystyle \,4\sqrt{x^2\,+\,y^2} \,+\,3x\;=\;5\)


Simplify: \(\displaystyle \,4\sqrt{x^2\,+\,y^2} \;=\;5\,-\,3x\)

Square: \(\displaystyle \,16(x^2\,+\,y^2) \;= \;(5\,-\,3x)^2\)

Expand: \(\displaystyle \,16x^2\,+\,16y^2\;=\;25\,-\,30x\,+\,9x^2\)

Therefore: \(\displaystyle \,7x^2\,+\,16y^2\,+\,30x\,-\,25\;=\;0\;\) . . . like you said: an ellipse

 

[fraiha]

Thank you for your help, I understand both of the problems now and tried some examples in the textbook and am sure i'll do well on my test. Thanks again!